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Total electric charge on the earth,

  1. Aug 23, 2010 #1
    1. The problem statement, all variables and given/known data
    The earth can be regarded as a sphere of radius R,
    the volume density of charge distributed as [tex] \rho = \frac{\rho_s r}{R} [/tex]
    where the density is 0 at thcce centre and rises linearly with radius until it reaches ps at the earth's surface

    i) prove that the total charge on the earth is [tex] \pi \rho_s R^3 [/tex]

    ii) use gauss's law to find an expression for E at the point r outside the earth (r>R)

    iii) use gauss's law to find an expression for E within the earths interior (r<R)

    iv) write expressions for the x-, y- and z- components of E at a point outside the earth and calculate [tex] \bf{\nabla .E }

    2. Relevant equations

    Qenclosed = [tex] \int_v \rho d\tau. [/tex] (for a volume)

    [tex] \oint_S \bf{E} . d\bf{a} = \frac{1}{\epsilon} Qenclosed [/tex]

    3. The attempt at a solution

    [tex] \int_v \rho \dtau [/tex]

    substituting in the given rho,

    [tex] \int \rho_s \frac{r}{R} \dtau[/tex]

    the outer radius r is = R so it just becomes R/R =1,

    [tex] \rho_s *1 \int \dtau [/tex]

    this is where I integrate the dtau and it should give me the volume of a sphere

    [tex] \rho_s \frac{4}{3}\pi r^3 [/tex]

    this sort of looks like the answer i'm looking for
    what have I done wrong?
    i've tried heaps of other methods, but I'm always getting some type of constant out the front
  2. jcsd
  3. Aug 23, 2010 #2


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    Science Advisor

    The density is proportional to r, as you wrote, but you can't pull this factor of r outside the integral. You should write:
    [tex]\int_0^R\rho(r)d\tau = \int_0^R\rho_s\frac{r}{R}4\pi r^2 dr = 4\pi\frac{\rho_s}{R}\int_0^Rr^3 dr = 4\pi \frac{\rho_s}{R} \frac{R^4}{4} = \pi\rho_sR^3[/tex]

    Does this make sense?
  4. Aug 23, 2010 #3
    Yes thanks, I actually ended up doing it like this myself last night, I just didn't update it here on PF
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